A electric charge question for two spheres

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SUMMARY

The discussion centers on the charge distribution between two metallic spheres, A and B, after they are connected by a wire. Sphere A, with a radius of 1.00 cm, initially holds a charge of 450 nC, while sphere B, with a radius of 2.00 cm, starts with no charge. Upon connecting the spheres, the charge redistributes, resulting in 150 nC on sphere A and 300 nC on sphere B, as confirmed by the equalization of electric potentials and the ratio of their surface areas. The correct answer is option D.

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Homework Statement



A metallic sphere A of radius 1.00 cm is several centimetres away from a metallic spherical shell B of radius 2.00 cm. Charge 450 nC is placed on A, with no charge on B or anywhere nearby. Next, the two objects are joined by a long, thin, metallic wire, and finally the wire is removed. How is the charge shared between A and B?
a. 225 nC on A and 225 nC on B.
b. 90.0 nC on A and 360 nC on B, with equal surface charge densities.
c. 0 on A, 450 nC on B.
d. 150 nC on A and 300 nC on B.
e. 450 nC on A and 0 on B.

Homework Equations





The Attempt at a Solution


The answer given is D..
My attempt was to determine the charge density when the two spheres are connected. Since they are both conudctors I took the fact that all the charge would be on the surface of both spheres. So dividing the charge Q by the total surface area of both spheres, gives a ratio of 1/4 of surface are for sphere A to sphere B. So 90/360, missing something here. thanks
 
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I think i worked this one out, when they are connected by the wire the potentials are equal so V1=V2, simplifys to
\frac{Q1}{Q2}=\frac{r1}{r2}=\frac{σ2}{σ1}
Leading to the correct answer..
 

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